Size of all connected non-empty cells of a Matrix

Given a binary matrix mat[][], the task is to find the size of all possible non-empty connected cells. 

An empty cell is denoted by 0 while a non-empty cell is denoted by 1. 
Two cells are said to be connected if they are adjacent to each other horizontally or vertically, i.e. mat[i][j] = mat[i][j – 1] or mat[i][j] = mat[i][j + 1] or mat[i][j] = mat[i – 1][j] or mat[i][j] = mat[i + 1][j]. 

Examples:

Input: mat[][] = {{1, 1, 0, 0, 0}, {0, 1, 0, 0, 1}, {1, 0, 0, 1, 1}, {0, 0, 0, 0, 0}, {1, 0, 1, 0, 1}} 
Output: 3 3 1 1 1 1 
Explanation:
{mat[0][0], mat[0][1], mat[1][1]}, {mat[1][4], mat[2][3], mat[2][4]}}, {mat[2][0]}, {mat[4][0]}, {mat[4][2]}, {mat[4[4]} are the only possible connections.

1 1 0 0 0

0 1 0 0 1

1 0 0 1 1

0 0 0 0 0

1 0 1 0 1

Input: mat[][] = {{1, 1, 0, 0, 0}, {1, 1, 0, 1, 1}, {1, 0, 0, 1, 1}, {0, 0, 0, 0, 0}, {0, 0, 1, 1, 1}} 
Output: 5 4 3 

Approach:
The idea is to use BFS and Recursion on the matrix. 
Follow the steps below: 

• Initialize a Queue Data Structure and insert a cell(mat[i][j] = 1).
• Perform BFS on the inserted cell and traverse its adjacent cells.
• Check for boundary conditions and check if the current element is 1, then flip it to 0.
• Mark the cell visited and update the size of the connected non-empty cells.
• Finally, print all the sizes of the connections obtained.

Below is the implementation of the above approach:

C++14

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find size of all the
// islands from the given matrix
int BFS(vector<vector<int> >& mat,
        int row, int col)
{
    int area = 0;

    // Initialize a queue for
    // the BFS traversal
    queue<pair<int, int> > Q;
    Q.push({ row, col });

    // Iterate until the
    // queue is empty
    while (!Q.empty()) {

        // Top element of queue
        auto it = Q.front();

        // Pop the element
        Q.pop();

        int r = it.first, c = it.second;

        // Check for boundaries
        if (r < 0 || c < 0 || r > 4 || c > 4)
            continue;

        // Check if current element is 0
        if (mat[r] == 0)
            continue;

        // Check if current element is 1
        if (mat[r] == 1) {

            // Mark the cell visited
            mat[r] = 0;

            // Incrementing the size
            area++;
        }

        // Traverse all neighbors
        Q.push({ r + 1, c });
        Q.push({ r - 1, c });
        Q.push({ r, c + 1 });
        Q.push({ r, c - 1 });
    }

    // Return the answer
    return area;
}

// Function to print size of each connections
void sizeOfConnections(vector<vector<int> > mat)
{

    // Stores the size of each
    // connected non-empty
    vector<int> result;

    for (int row = 0; row < 5; ++row) {
        for (int col = 0; col < 5; ++col) {

            // Check if the cell is
            // non-empty
            if (mat[row][col] == 1) {

                // Function call
                int area = BFS(mat, row, col);
                result.push_back(area);
            }
        }
    }

    // Print the answer
    for (int val : result)
        cout << val << " ";
}

// Driver Code
int main()
{

    vector<vector<int> > mat
        = { { 1, 1, 0, 0, 0 },
            { 1, 1, 0, 1, 1 },
            { 1, 0, 0, 1, 1 },
            { 1, 0, 0, 0, 0 },
            { 0, 0, 1, 1, 1 } };

    sizeOfConnections(mat);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;

class GFG{

static class pair
{
    int first, second;
    pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find size of all the
// islands from the given matrix
static int BFS(int[][] mat,
               int row, int col)
{
    int area = 0;

    // Initialize a queue for
    // the BFS traversal
    Queue<pair> Q = new LinkedList<>();
    Q.add(new pair(row, col));

    // Iterate until the
    // queue is empty
    while (!Q.isEmpty())
    {

        // Top element of queue
        pair it = Q.peek();

        // Pop the element
        Q.poll();

        int r = it.first, c = it.second;

        // Check for boundaries
        if (r < 0 || c < 0 ||
            r > 4 || c > 4)
            continue;

        // Check if current element is 0
        if (mat[r] == 0)
            continue;

        // Check if current element is 1
        if (mat[r] == 1)
        {

            // Mark the cell visited
            mat[r] = 0;

            // Incrementing the size
            area++;
        }

        // Traverse all neighbors
        Q.add(new pair(r + 1, c));
        Q.add(new pair(r - 1, c));
        Q.add(new pair(r, c + 1));
        Q.add(new pair(r, c - 1));
    }

    // Return the answer
    return area;
}

// Function to print size of each connections
static void sizeOfConnections(int[][] mat)
{

    // Stores the size of each
    // connected non-empty
    ArrayList<Integer> result = new ArrayList<>();

    for(int row = 0; row < 5; ++row)
    {
        for(int col = 0; col < 5; ++col)
        {

            // Check if the cell is
            // non-empty
            if (mat[row][col] == 1)
            {

                // Function call
                int area = BFS(mat, row, col);
                result.add(area);
            }
        }
    }

    // Print the answer
    for(int val : result)
       System.out.print(val + " ");
}

// Driver code
public static void main (String[] args)
{
    int[][] mat = { { 1, 1, 0, 0, 0 },
                    { 1, 1, 0, 1, 1 },
                    { 1, 0, 0, 1, 1 },
                    { 1, 0, 0, 0, 0 },
                    { 0, 0, 1, 1, 1 } };

    sizeOfConnections(mat);
}
}

// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
from collections import deque

# Function to find size of all the
# islands from the given matrix
def BFS(mat, row, col):

    area = 0

    # Initialize a queue for
    # the BFS traversal
    Q = deque()
    Q.append([row, col])

    # Iterate until the
    # queue is empty
    while (len(Q) > 0):

        # Top element of queue
        it = Q.popleft()

        # Pop the element
        # Q.pop();
        r, c = it[0], it[1]

        # Check for boundaries
        if (r < 0 or c < 0 or r > 4 or c > 4):
            continue

        # Check if current element is 0
        if (mat[r] == 0):
            continue

        # Check if current element is 1
        if (mat[r] == 1):

            # Mark the cell visited
            mat[r] = 0

            # Incrementing the size
            area += 1

        # Traverse all neighbors
        Q.append([r + 1, c])
        Q.append([r - 1, c])
        Q.append([r, c + 1])
        Q.append([r, c - 1])

    # Return the answer
    return area

# Function to prsize of each connections
def sizeOfConnections(mat):

    # Stores the size of each
    # connected non-empty
    result = []

    for row in range(5):
        for col in range(5):

            # Check if the cell is
            # non-empty
            if (mat[row][col] == 1):

                # Function call
                area = BFS(mat, row, col);
                result.append(area)

    # Print the answer
    for val in result:
        print(val, end = " ")

# Driver Code
if __name__ == '__main__':

    mat = [ [ 1, 1, 0, 0, 0 ],
            [ 1, 1, 0, 1, 1 ],
            [ 1, 0, 0, 1, 1 ],
            [ 1, 0, 0, 0, 0 ],
            [ 0, 0, 1, 1, 1 ] ]

    sizeOfConnections(mat)

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

class pair
{
    public int first, second;

    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find size of all the
// islands from the given matrix
static int BFS(int[, ] mat, int row, int col)
{
    int area = 0;

    // Initialize a queue for
    // the BFS traversal
    Queue<pair> Q = new Queue<pair>();
    Q.Enqueue(new pair(row, col));

    // Iterate until the
    // queue is empty
    while (Q.Count != 0)
    {

        // Top element of queue
        pair it = Q.Peek();

        // Pop the element
        Q.Dequeue();

        int r = it.first, c = it.second;

        // Check for boundaries
        if (r < 0 || c < 0 || r > 4 || c > 4)
            continue;

        // Check if current element is 0
        if (mat[r, c] == 0)
            continue;

        // Check if current element is 1
        if (mat[r, c] == 1)
        {

            // Mark the cell visited
            mat[r, c] = 0;

            // Incrementing the size
            area++;
        }

        // Traverse all neighbors
        Q.Enqueue(new pair(r + 1, c));
        Q.Enqueue(new pair(r - 1, c));
        Q.Enqueue(new pair(r, c + 1));
        Q.Enqueue(new pair(r, c - 1));
    }

    // Return the answer
    return area;
}

// Function to print size of each connections
static void sizeOfConnections(int[,] mat)
{

    // Stores the size of each
    // connected non-empty
    ArrayList result = new ArrayList();

    for(int row = 0; row < 5; ++row)
    {
        for(int col = 0; col < 5; ++col)
        {

            // Check if the cell is
            // non-empty
            if (mat[row, col] == 1)
            {

                // Function call
                int area = BFS(mat, row, col);
                result.Add(area);
            }
        }
    }

    // Print the answer
    foreach(int val in result)
        Console.Write(val + " ");
}

// Driver code
public static void Main(string[] args)
{
    int[, ] mat = { { 1, 1, 0, 0, 0 },
                    { 1, 1, 0, 1, 1 },
                    { 1, 0, 0, 1, 1 },
                    { 1, 0, 0, 0, 0 },
                    { 0, 0, 1, 1, 1 } };

    sizeOfConnections(mat);
}
}

// This code is contributed by grand_master

Javascript

<script>

// Javascript program to implement
// the above approach

class pair
{
    constructor(first, second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find size of all the
// islands from the given matrix
function BFS(mat, row, col)
{
    var area = 0;

    // Initialize a queue for
    // the BFS traversal
    var Q = [];
    Q.push(new pair(row, col));

    // Iterate until the
    // queue is empty
    while (Q.length != 0)
    {

        // Top element of queue
        var it = Q[0];

        // Pop the element
        Q.shift();

        var r = it.first, c = it.second;

        // Check for boundaries
        if (r < 0 || c < 0 || r > 4 || c > 4)
            continue;

        // Check if current element is 0
        if (mat[r][ c] == 0)
            continue;

        // Check if current element is 1
        if (mat[r] == 1)
        {

            // Mark the cell visited
            mat[r] = 0;

            // Incrementing the size
            area++;
        }

        // Traverse all neighbors
        Q.push(new pair(r + 1, c));
        Q.push(new pair(r - 1, c));
        Q.push(new pair(r, c + 1));
        Q.push(new pair(r, c - 1));
    }

    // Return the answer
    return area;
}

// Function to print size of each connections
function sizeOfConnections(mat)
{

    // Stores the size of each
    // connected non-empty
    var result = [];

    for(var row = 0; row < 5; ++row)
    {
        for(var col = 0; col < 5; ++col)
        {

            // Check if the cell is
            // non-empty
            if (mat[row][col] == 1)
            {

                // Function call
                var area = BFS(mat, row, col);
                result.push(area);
            }
        }
    }

    // Print the answer
    for(var val of result)
        document.write(val + " ");
}

// Driver code
var mat = [ [ 1, 1, 0, 0, 0 ],
                [ 1, 1, 0, 1, 1 ],
                [ 1, 0, 0, 1, 1 ],
                [ 1, 0, 0, 0, 0 ],
                [ 0, 0, 1, 1, 1 ] ];
sizeOfConnections(mat);


</script>

Output:

6 4 3

Time Complexity: O(row * col)
Auxiliary Space: O(row * col)

Example in c/c++:

Approach:

Define a utility function is_valid to check if a given cell (i, j) is a valid cell in the matrix.

Define a recursive function dfs to perform DFS on the matrix and count the size of each connected component. The function takes the following arguments:

matrix: the 2D matrix to be traversed
visited: a 2D matrix to keep track of visited cells
i: the row index of the current cell
j: the column index of the current cell
size: a pointer to an integer to keep track of the size of the current connected component
The function starts by marking the current cell as visited and incrementing the size of the current component. Then, it checks all adjacent cells and recursively calls the dfs function on unvisited neighbors.

In the main function, define a 2D matrix matrix to be traversed, and initialize all cells in a 2D matrix visited as unvisited.

Traverse the matrix and count the size of each connected component starting from each non-empty, unvisited cell. Print the size of each connected component to the console.

C++

#include<bits/stdc++.h>
using namespace std;

#define ROWS 4
#define COLS 4

// A utility function to check if a given cell (i, j) is a valid cell in the matrix
int is_valid(int i, int j) {
    return (i >= 0 && i < ROWS && j >= 0 && j < COLS);
}

// A recursive function to perform DFS on the matrix and count the size of each connected component
void dfs(vector<vector<int>> &matrix, vector<vector<int>> &visited, int i, int j, int* size) {
    // Mark the current cell as visited and increment the size of the current component
    visited[i][j] = 1;
    (*size)++;

    // Check all adjacent cells and recursively call the dfs function on unvisited neighbors
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, 1, 0, -1};
    for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
        if (is_valid(x, y) && matrix[x][y] && !visited[x][y]) {
            dfs(matrix, visited, x, y, size);
        }
    }
}

int main() {
    vector<vector<int>> matrix= {
        {1, 0, 1, 0},
        {1, 1, 0, 1},
        {0, 1, 0, 0},
        {1, 1, 1, 1}
    };
    vector<vector<int>> visited(ROWS, vector<int>(COLS, 0)); // Initialize all cells as unvisited
    int component_size = 0;

    // Traverse the matrix and count the size of each connected component
    for (int i = 0; i < ROWS; i++) {
        for (int j = 0; j < COLS; j++) {
            if (matrix[i][j] && !visited[i][j]) {
                component_size = 0;
                dfs(matrix, visited, i, j, &component_size);
                cout << "Size of connected component starting at ("<< i << ", " << j  <<") is " << component_size << endl;
            }
        }
    }
    return 0;
}

// The code is contributed by Arushi Jindal.

C

#include <stdio.h>

#define ROWS 4
#define COLS 4

// A utility function to check if a given cell (i, j) is a valid cell in the matrix
int is_valid(int i, int j) {
    return (i >= 0 && i < ROWS && j >= 0 && j < COLS);
}

// A recursive function to perform DFS on the matrix and count the size of each connected component
void dfs(int matrix[][COLS], int visited[][COLS], int i, int j, int* size) {
    // Mark the current cell as visited and increment the size of the current component
    visited[i][j] = 1;
    (*size)++;

    // Check all adjacent cells and recursively call the dfs function on unvisited neighbors
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, 1, 0, -1};
    for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
        if (is_valid(x, y) && matrix[x][y] && !visited[x][y]) {
            dfs(matrix, visited, x, y, size);
        }
    }
}

int main() {
    int matrix[ROWS][COLS] = {
        {1, 0, 1, 0},
        {1, 1, 0, 1},
        {0, 1, 0, 0},
        {1, 1, 1, 1}
    };
    int visited[ROWS][COLS] = {0}; // Initialize all cells as unvisited
    int component_size = 0;

    // Traverse the matrix and count the size of each connected component
    for (int i = 0; i < ROWS; i++) {
        for (int j = 0; j < COLS; j++) {
            if (matrix[i][j] && !visited[i][j]) {
                component_size = 0;
                dfs(matrix, visited, i, j, &component_size);
                printf("Size of connected component starting at (%d, %d) is %d\n", i, j, component_size);
            }
        }
    }
    return 0;
}

Python3

ROWS = 4
COLS = 4

# A utility function to check if a given cell (i, j) is a valid cell in the matrix
def is_valid(i, j):
    return (i >= 0 and i < ROWS and j >= 0 and j < COLS)

# A recursive function to perform DFS on the matrix and count the size of each connected component
def dfs(matrix, visited, i, j, size):
    # Mark the current cell as visited and increment the size of the current component
    visited[i][j] = 1
    size[0] += 1

    # Check all adjacent cells and recursively call the dfs function on unvisited neighbors
    dx = [-1, 0, 1, 0]
    dy = [0, 1, 0, -1]
    for k in range(4):
        x = i + dx[k]
        y = j + dy[k]
        if is_valid(x, y) and matrix[x][y] and not visited[x][y]:
            dfs(matrix, visited, x, y, size)

matrix= [
    [1, 0, 1, 0],
    [1, 1, 0, 1],
    [0, 1, 0, 0],
    [1, 1, 1, 1]
]

visited = [[0]*COLS for _ in range(ROWS)] # Initialize all cells as unvisited
component_size = [0]

# Traverse the matrix and count the size of each connected component
for i in range(ROWS):
    for j in range(COLS):
        if matrix[i][j] and not visited[i][j]:
            component_size = [0]
            dfs(matrix, visited, i, j, component_size)
            print(f"Size of connected component starting at ({i}, {j}) is {component_size[0]}")

# This code is contributed by phasing17

Output

Size of connected component starting at (0, 0) is 8
Size of connected component starting at (0, 2) is 1
Size of connected component starting at (1, 3) is 1

The time complexity of this approach is O(rows * cols), as we need to visit every cell in the matrix once. 

The space complexity is O(rows * cols), as we need to create a separate visited array to keep track of which cells have already been visited.